In a Young's double slit experiment, if the slit separation is twice the wavelength of light used, then the…
In a Young's double slit experiment, if the slit separation is twice the wavelength of light used, then the maximum number of interference maxima is
0
3
5
7
Solution
For possible interference maxima on the screen the condition is
\(\begin{aligned}
d \sin \theta & =n \lambda \quad \ldots (i) \\
\text{Here, } d & =2 \lambda \text { (Given) } \\
2 \lambda \sin \theta & =n \lambda \\
2 \sin \theta & =n
\end{aligned}\)
The maximum value of \(\sin \theta\) is 1 .
Hence,
\(n=2 \times 1=2\)
Thus the Eq. (i) must be satisfied by 5 integer values i.e. \(-2,-1,0,1,2\).
Hence, the maximum number of possible interference maxima is 5 .